# 2001 IMO Shortlist Problems/C3

## Contents

## Problem

Define a -*clique* to be a set of people such that every pair of them are acquainted with each other. At a certain party, every pair of 3-cliques has at least one person in common, and there are no 5-cliques. Prove that there are two or fewer people at the party whose departure leaves no 3-clique remaining.

## Solution

### Solution 1

If there exists only one 3-clique, remove anyone in that clique. (If there are no 3-cliques, we are done!) Otherwise, consider the following cases:

**Case 1: There exist a pair of 3-cliques that share 2 people.**

Let these 3-cliques be and . If every other 3-clique contained either or , then removing and will leave no 3-clique remaining.

Otherwise, if there was a 3-clique that did not contain or , then it would have to contain and to satisfy the condition. Call this 3-clique . First notice that and are also 3-cliques now since and are now acquainted. We claim that removing and will now suffice; suppose there was another 3-clique not containing or . This 3-clique must contain , , and (to share with , , and ). However, this means is a 5-clique; contradiction.

**Case 2: Every pair of 3-cliques has at most one person in common.**

There exist two 3-cliques that share a person whom we will call : name the 3-cliques and . We now claim that every 3-clique in this case must contain .

Suppose there existed a 3-clique not containing . It must share one person with each of and ; without loss of generality, let this 3-clique contain and . But this means and are acquainted, and due to their mutual acquaintance with , is a 3-clique, which shares 2 people with , contradicting the premise of this case.

Thus, every 3-clique contains , and so removing will remove every 3-clique.